Given W is a uniformly distributed random variable with mean 33 and variance 3. determine:
(a) probability density function for W
(b) cumulative distribution function for W
(a) probability density function for W
The probability distribution function of a uniform distribution is defined as below
f(w) = 1/ (b-a) for a≤ w ≤ b
The mean = ( a + b)/ (3) = 33 for our case.
The variance = (b-a)2/12 = 3 for our case.
solving the two equations above to obtain a and b we have
a + b = 99 (first equation)
(b - a)2 = 36 which becomes (b-a) = 6 which becomes b = a + 6 (second equation)
substituting the second equation in the first equation, we have
a + a + 6 = 99
2a = 93
a = 46.5
Thus b = (46.5 + 6) = 52.5
substituting b = 52.5, a = 46.5 in f(x) = 1/ (b-a) for a≤ w ≤ b, we obtain
f(w) = 1/6 for 46.5 ≤ w ≤ 52.5
0, otherwise
which is the required probability distribution of w
(b) cumulative distribution function for W
we integrate the probability distribution function of w to obtain the cumulative distribution function for W .
we obtain as below
"\\smallint" 46.5w (1/6) dw
integrating we obtain (1/6)w46.5 = ( (1/6)w - 7.75 )
Thus , the cumulative distribution of w is F(w) = ( (1/6)w - 7.75 ) 46.5 ≤ w ≤ 52.5
0, otherwise
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